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A rod (length = l) is rotating about it s center with a speed 'w' at temperature to . If the temperature is increased to $\;t_{0}+\bigtriangleup t\;$ , the new speed of rotation is -

$(a)\;w(1+2 \alpha \bigtriangleup t)\qquad(b)\;w(1+ \alpha \bigtriangleup t)\qquad(c)\;\large\frac{w}{w(1+2 \alpha \bigtriangleup t)}\qquad(d)\;\large\frac{w}{w(1+ \alpha \bigtriangleup t)}$

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Answer : $\large\frac{w}{w(1+2 \alpha \bigtriangleup t)}$
Explanation :
Since the angular momentum is conserved ,
$\large\frac{ml^2}{12}w=\large\frac{m(l+\bigtriangleup l)^2}{12} w_{2}$
$\large\frac{ml^2}{12}w=\large\frac{ml^2}{12}(1+\large\frac{\bigtriangleup l}{l})^2 w_{2}$
$w \approx (1+2 \large\frac{2 \bigtriangleup l}{l})w_{2}$
$[\large\frac{\bigtriangleup l}{l} \to 0]$
$w \approx (1 + 2 \alpha \bigtriangleup t) w_{2} $
$w_{2} \approx \large\frac{w}{(1+2 \alpha \bigtriangleup)}$
answered Mar 13, 2014 by yamini.v

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